Apoliceman, parked at arn intersection, is passed by a speeder traveling at 120 km/hr. two seconds later, the policeman starts and uniformly accelerates at 4.0 m/s^2. how far will the policeman travel before he catches the speeder?

682.32 m

Explanation:

Speed of the passing speeder = 120 km/hr = 120 × 0.2777 = 33.33 m/s

time after which police man starts = 2 seconds

Acceleration of the policeman = 4.0 m/s²

let the time taken to catch be 't' seconds

Now,

The total distance to be covered by the policeman

= Distance covered by the speeder in the 2 seconds + Distance further traveled by the speeder in time t

thus,

From Newton's equation of motion

where,  

s is the total distance traveled by the police man

u is the initial speed  = 0

a is the acceleration

t is the time

thus,

= 33.33 × 2 + 33.33 × t

or

2t² = 66.66 + 33.33t

or

2t² - 33.33t - 66.66 = 0

on solving the above equation, we get

t = 18.47 seconds (negative value is ignored as time cannot be negative)

therefore,

the total distance covered = 33.33 × 2 + 33.33 × 18.47 = 682.32 m

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form